Question

Prove the identity: SinA CosB+ CosA SinB/SinA+ sinB= sinA- SinB/ sinA cosB - cosA sinB

Answer 1

Answer:

[math]E = \frac {\cos(A) - \cos(B)}{\sin(A) + \sin(B)}[/math]

Multiplying the numerator and denominator by [math]\cos(A) + \cos(B)[/math]

[math]E = \frac {\cos(A) - \cos(B)}{\sin(A) + \sin(B)} \times \frac {\cos(A) + \cos(B)}{\cos(A) + \cos(B)}[/math]

[math]= \frac {\cos^2(A) - \cos^2(B)}{[\sin(A) + \sin(B)][\cos(A) + \cos(B)]}[/math]

Using the identities [math]\cos^2(A) = 1 - \sin^2(A)[/math] and [math]\cos^2(B) = 1 - \sin^2(B)[/math]

[math]E = \frac {1 - \sin^2(A) - 1 + \sin^2(B)}{[\sin(A) + \sin(B)][\cos(A) + \cos(B)]}[/math]

[math]= \frac {\sin^2(B) - \sin^2(A)}{[\sin(A) + \sin(B)][\cos(A) + \cos(B)]}[/math]

[math]= \frac {[\sin(B) - \sin(A)][\sin(B) + \sin(A)]}{[\sin(A) + \sin(B)][\cos(A) + \cos(B)]}[/math]

Dividing the numerator and denominator by [math]\sin(A) + \sin(B)[/math], we have:

[math]E = \frac {\sin(B) - \sin(A)}{\cos(A) + \cos(B)}[/math]

Answer 2

Answer:

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